$$ \Vert uv\Vert_{H^m} \leq C\sum_{j=2}^m \Vert u\Vert_{H_j} \Vert v\Vert_{H^{m-j+2}}, m\geq 2$$ This inequality from
Auzinger, Winfried; Kassebacher, Thomas; Koch, Othmar; Thalhammer, Mechthild, Convergence of a strang splitting finite element discretization for the Schrödinger-Poisson equation, ESAIM, Math. Model. Numer. Anal. 51, No. 4, 1245-1278 (2017). ZBL1379.65071.
This paper said
By the Hölder inequality and the Sobolev embeddings of $H^2$ in $L^\infty$ and $H^1$ in $L^4$ , we have$$ \Vert uv\Vert_{H^m} \leq C\sum_{j=2}^m \Vert u\Vert_{H_j} \Vert v\Vert_{H^{m-j+2}}, m\geq 2$$
I don't konw how to prove this inequality.
By Leibniz formula, $\nabla^n(uv)$ will give you terms of the form $\nabla^k u \nabla^{n-k}v$ with $0\leq k\leq n$. Then if $k<m$ and $n-k< m$, you use the following Holder inequality $$ \|\nabla^k u \nabla^{n-k}v\|_{L^2} \leq \|\nabla^ku\|_{L^4} \|\nabla^{n-k}v\|_{L^4} $$ and so the mentioned Sobolev embedding gives $$ \|\nabla^k u \nabla^{n-k}v\|_{L^2} \leq \|u\|_{H^{k+1}} \|v\|_{H^{n-k+1}} $$ which is in the range of the sum. If $n-k=m$, so $k=0$ and $n=m$, then you do not want to get any $H^{m+1}$ appearing, so you write instead $$ \|u \nabla^{m}v\|_{L^2} \leq \|u\|_{L^\infty} \|\nabla^mv\|_{L^2} $$ and so the other mentioned Sobolev embedding gives $$ \| u \nabla^{m}v\|_{L^2} \leq \|u\|_{H^{2}} \|v\|_{H^{m}} $$ which again is bounded by the sum. The same applies to the case $k=n=m$.